A sequence is defined by the recurrence relation $a_n = 11a_{n-1} + 3n$ for $n \geq 1$, with initial condition $a_0 = 1$. Find the smallest positive integer $n$ such that the sum of the first $n$ terms of the sequence is divisible by 11.

To solve the problem, let’s break it down step-by-step. We need to find the **smallest positive integer $ n $** such that the **sum of the first $ n $** terms of the sequence defined by the recurrence relation 

\[
a_n = 11a_{n-1} + 3n, \quad a_0 = 1
\]

is divisible by 11. 

---

### Step 1: Calculate a few terms of the sequence $ a_n $
We start with the initial condition $ a_0 = 1 $ and use the recurrence relation to compute the next few terms to understand the sequence better.

- **For $ n = 1 $:**
  \[
  a_1 = 11a_0 + 3(1) = 11 \cdot 1 + 3 = 14
  \]

- **For $ n = 2 $:**
  \[
  a_2 = 11a_1 + 3(2) = 11 \cdot 14 + 6 = 154 + 6 = 160
  \]

- **For $ n = 3 $:**
  \[
  a_3 = 11a_2 + 3(3) = 11 \cdot 160 + 9 = 1760 + 9 = 1769
  \]

So, the first few terms are:
\[
a_0 = 1, \quad a_1 = 14, \quad a_2 = 160, \quad a_3 = 1769, \quad \dots
\]

---

### Step 2: Define the Sum $ S_n $
We are asked to find the smallest $ n $ such that the **sum of the first $ n $** terms is divisible by 11. Let

\[
S_n = a_0 + a_1 + a_2 + \dots + a_n.
\]

We need to find the **smallest positive $ n $** such that $ S_n \equiv 0 \pmod{11} $.

---

### Step 3: Compute the Partial Sums
Let’s compute the partial sums step-by-step and reduce them modulo 11 to see when the divisibility condition is satisfied.

- **$ S_1 = a_0 + a_1 = 1 + 14 = 15 $**  
  \[ 15 \equiv 4 \pmod{11}. \]

- **$ S_2 = S_1 + a_2 = 15 + 160 = 175 $**  
  \[ 175 \equiv 10 \pmod{11}. \]

- **$ S_3 = S_2 + a_3 = 175 + 1769 = 1944 $**  
  \[ 1944 \equiv 7 \pmod{11}. \]

Let’s continue further.

- **$ S_4 = S_3 + a_4 $** (we need $ a_4 $):
  \[
  a_4 = 11a_3 + 3(4) = 11 \cdot 1769 + 12 = 19459 + 12 = 19471.
  \]
  Thus:
  \[ S_4 = 1944 + 19471 = 21415. \]
  \[ 21415 \equiv 3 \pmod{11}. \]

- **$ S_5 = S_4 + a_5 $** (we need $ a_5 $):
  \[
  a_5 = 11a_4 + 3(5) = 11 \cdot 19471 + 15 = 214181 + 15 = 214196.
  \]
  Thus:
  \[ S_5 = 21415 + 214196 = 235611. \]
  \[ 235611 \equiv 0 \pmod{11}. \]

---

### Step 4: Conclusion
The smallest positive integer $ n $ such that the sum of the first $ n $ terms is divisible by 11 is:

\[
\boxed{5}.
\]

---

Would you like further details or clarifications? 

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Here are 5 related questions for deeper exploration:

1. How can we find a closed formula for the sequence $ a_n $?
2. What are the properties of sequences defined by linear recurrence relations?
3. How can modular arithmetic be used more systematically in recurrence problems?
4. Can generating functions help solve such recurrence relations?
5. What is the asymptotic behavior of this sequence $ a_n $ as $ n \to \infty $?

**Tip:** When working with recurrences, computing a few initial terms often gives crucial insights into patterns and helps with verification.

Solve a recurrence relation problem where a_n = 11a_(n-1) + 3n with initial condition a_0 = 1. Learn how to compute the sum of the sequence's terms and find the smallest n such that the sum is divisible by 11 using modular arithmetic. Ideal for students in Grades 10-12.

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